Substitution Practice √ Use a trigonometric substitution to integrate the function f (x) = x x2 − 9. Check your work by integration using the substitution u = x2 . Solution Referring to our trig substitution summary, we see that the recommended way √ to integrate an expression including x2 − 32 is to substitute x = 3 sec θ, in which case dx = 3 sec θ tan θ dθ and: � � x2 − 9 = (3 sec θ)2 − 9 � = 9 sec2 θ − 9 � = 3 sec2 θ − 1 � x2 − 9 = 3 tan θ. We start by performing this substitution and simplifying: � � � 2 x x − 9 dx = (3 sec θ)(3 tan θ)3 sec θ tan θ dθ � = 27 sec2 θ tan2 θ dθ. At this point we substitute u = tan θ, so du = sec2 θ dθ and: � � � x x2 − 9 dx = 27 sec2 θ tan2 θ dθ � = 27 u2 du = = u3 +c 3 9 tan3 θ + c 27 We have an answer in terms of tan θ and we want an answer in terms of x. We know that x = 3 sec θ, or equivalently that sec θ = x3 . Keeping that fact in mind (or the fact that cos θ = x3 ) we draw a right triangle with one angle equal to θ in which the side adjacent to θ has length 3 and the hypotenuse has length x. By the Pythagorean theorem, the side opposite the angle θ has length √ x2 − 9 and: √ x2 − 9 tan θ = . 3 We can now complete our calculation: � � x x2 − 9 dx = 9 tan3 θ + c � x � x2 − 9 dx = �√ �3 x2 − 9 9 +c 3 = 1 2 (x − 9)3/2 + c 3 While the calculation above is correct, it is faster and more reliable to com pute this integral via the substitution u = x2 − 9. If we do this we have du = 2x dx or x dx = 12 du and: � � � √ 1 2 x x − 9 dx = u du 2 1 2 3/2 = · u +c 2 3 1 3/2 = u +c 3 � � 1 2 x x2 − 9 dx = (x − 9)3/2 + c. 3

Substitution Practice

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